as part of an audio project I've been working on, I've been measuring the current consumption of a class A amplifier driving a 4Ohm 5W speaker (square wave signal).
Strangely, I notice that this:
Draws more current than this:
Why is that?
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1\$\begingroup\$ Hint: Your 1st circuit is an amplifier and your 2nd circuit is a voltage follower. \$\endgroup\$– brhansCommented Jun 14, 2018 at 18:41
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2\$\begingroup\$ Hint 2: Your first circuit can have voltage gain while your second circuit cannot, as the speaker interferes with voltage gain, but not current gain. \$\endgroup\$– user105652Commented Jun 14, 2018 at 18:46
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\$\begingroup\$ Hint 3: See if your observation holds for all input volume levels. \$\endgroup\$– Phil FrostCommented Jun 14, 2018 at 23:42
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2 Answers
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In your first circuit, the signal is voltage amplified by Vin * hFE of the transistor, until you reach a level that approaches Vdd to ground, causing distortion. Vbe can never be more than 0.7 volts.
In your second design there is no voltage amplification but current of the signal is boosted. It will appear at the speaker terminals minus 0.7 volts, the Vbe drop. Output volume will be much less, but saturate quicker as the Vbe drop subtracts from your voltage swing across the speaker.
Note that if your signal source is relatively high-impedance, as in a volume control, both designs will give you a boost in sound level. Both designs take advantage of the transistors hFE. The first design you can saturate the transistor fully ON, so more current is consumed.
In the second design the speaker will not let the emitter go to zero volts, even with a strong signal. This mode will cause distortion sooner as you have a voltage range of just ground to Vdd -0.7 volts minus the Vce drop. Putting the speaker on the emitter creates some negative feedback, further reducing the gain.
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\$\begingroup\$ Your 4th paragraph cleared the cobwebs off ;-) \$\endgroup\$– SteveCommented Oct 8, 2019 at 23:17
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The first circuit will saturate to within about 0.2V of ground, so your voltage across the speaker is VDD-0.2V. The second, if the base is driven as high as VDD, will have a best case output level of VDD-0.7V. It could be lower if the input is lower than VDD or your base resistor is too high.
answered Jun 14, 2018 at 19:24